The Three-Moment Equation for
Continuous-Beam Analysis
CEE 201L. Uncertainty, Design, and Optimization
Department of Civil and Environmental Engineering
Duke University
Henri P. Gavin
Spring, 2009
Consider a continuous beam over several supports carrying arbitrary loads,
w(x).
Using the Moment-Area Theorem, we will analyze two adjoining spans of this
beam to find the relationship between the internal moments at each of the
supports and the loads applied to the beam. We will label the left, center,
and right supports of this two-span segment L, C, and R. The left span has
length LL and flexural rigidity EIL ; the right span has length LR and flexural
rigidity EIR (see figure (a)).
Applying the principle of superposition to this two-span segment, we can
separate the moments caused by the applied loads from the internal moments
at the supports. The two-span segment can be represented by two simplysupported spans (with zero moment at L, C, and R) carrying the external
loads plus two simply-supported spans carrying the internal moments ML ,
MC , and MR (figures (b), (c), and (d)). The applied loads are illustrated
below the beam, so as not to confuse the loads with the moment diagram
(shown above the beams). Note that we are being consistent with our sign
convention: positive moments create positive curvature in the beam. The
internal moments ML , MC , and MR are drawn in the positive directions. The
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CEE 201L. – Uncertainty, Design, and Optimization – Duke University – Spring 2009 – H.P. Gavin
areas under the moment diagrams due to the applied loads on the simplysupported spans (figure (b)) are AL and AR ; x̄L represents the distance from
the left support to the centroid of AL , and x̄R represents the distance from
the right support to the centroid of AR , as shown. The moment diagrams
due to the unknown moments, ML , MC , and MR are triangular, as shown in
figures (c) and (d).
Examining the elastic curve of the continuous beam (figure (e)), we recognize
that the rotation of the beam at the center support, θC , is continuous across
support C. In other words, θC just to the left of point C is the same as θC
just to the right of point C. This continuity condition may be expressed
∆R tan C
∆L tan C
=−
,
(1)
LL
LR
where ∆L tan C is the distance from the tangent at C to point L, and ∆R tan C
is the distance from the tangent at C to point R.
Using the second Moment-Area Theorem, and assuming that the flexural
rigidity (EI) is constant within each span, we can find the terms ∆L tan C ,
and ∆R tan C in terms of the unknown moments, ML , MC , and MR and the
known applied loads.
"
#
1
2 1
1 1
∆L tan C =
x̄L AL + LL MC LL + LL ML LL ,
(2)
EIL
3 2
3 2
and
#
"
1
2 1
1 1
∆R tan C =
x̄R AR + LR MC LR + LR MR LR .
(3)
EIR
3 2
3 2
Substituting these expressions into equation (1) and re-arranging terms, leads
to the three-moment equation. 1
LL
LL
LR
LR
6x̄L AL
6x̄R AR
ML + 2
+
MC +
MR = −
−
. (4)
EIL
EIL EIR
EIR
LL EIL LR EIR
Note that if EIL = EIR = EI, the three-moment equation is independent of
!
EI.
1
The three-moment equation was derived by Émile Clapeyron in 1857 using the differential equations of
beam bending.
CC BY-NC-ND H.P. Gavin
3
Strain Energy in Linear Elastic Solids
w(x)
(a)
111
000
000
111
000
111
111
000
000
111
000
111
EIL
L
111
000
000
111
000
111
R
EIR
C
LL
=
LR
xL
xR
AL
(b)
111
000
000
111
000
111
AR
w(x)
+
111
000
000
111
000
111
111
000
000
111
2
3 LL
2
3 LR
(c)
111
000
000
111
000
111
000
M111
000M
111
111
000
000
111
000
111
000
111
C
C
1L
3 L
+
1L
3 R
(d)
M111
000
L
M
111
000
000 R
111
111
000
000
111
000
111
θC
(e)
11
00
00
11
∆ L tan C
θC
111
000
000
111
∆ R tan C
11
00
00
11
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4
CEE 201L. – Uncertainty, Design, and Optimization – Duke University – Spring 2009 – H.P. Gavin
To apply the three-moment equation numerically, the lengths, moments of
inertia, and applied loads must be specified for each span. Two commonly
applied loads are point loads and uniformly distributed loads. For point loads
PL and PR acting a distance xL and xR from the left and right supports,
respectively, the right hand side of the three-moment equation becomes
−
6x̄R AR
xL
xR
6x̄L AL
−
= −PL
(L2L − x2L ) − PR
(L2R − x2R ) . (5)
LL EIL LR EIR
LL EIL
LR EIR
For uniformly distributed loads wL and wR on the left and right spans,
6x̄L AL
6x̄R AR
wL L3L wR L3R
−
−
=−
−
.
LL EIL LR EIR
4EIL
4EIR
(6)
To find the internal moments at the N + 1 supports in a continuous beam
with N spans, the three-moment equation is applied to N − 1 adjacent pairs
of spans. For example, consider the application of the three-moment equation
to a four-span beam. Spans a, b, c, and d carry uniformly distributed loads
wa , wb , wc , and wd , and rest on supports 1, 2, 3, 4, and 5.
The three-moment equation is applied to spans a − b, spans b − c, and spans
c − d.
La
M1 + 2
EIa
Lb
M2 + 2
EIb
Lc
M3 + 2
EIc
La
Lb
Lb
wa L3a wb L3b
+
M2 +
M3 = −
−
EIa EIb
EIb
4EIa 4EIb
!
Lb
Lc
Lc
wb L3b wc L3c
+
M4 = −
−
M3 +
EIb EIc
EIc
4EIb 4EIc
!
Lc
Ld
Ld
wc L3c wd L3d
+
M4 +
M5 = −
−
EIc EId
EId
4EIc 4EId
!
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5
Strain Energy in Linear Elastic Solids
We know that M1 = 0 and M5 = 0 because they are at the ends of the span.
Applying these end-moment conditions to the three three-moment equations
and casting the equations into matrix form,

1





 La

 EIa






 0






 0






0
0
2
La
EIa
+
Lb
EIb
0
Lb
EIb
0
Lb
EIb
2
Lb
EIb
+
0
Lc
EIc
0
0
Lc
EIc
2
0
0
0
Lc
EIc
0
Lc
EIc
+
Ld
EId
Ld
EId
0
1





































M2 





M3 





M4 




M5



























M1
=
3
wb L3b
4EIb
3
wc L3c
4EIc
3
wd L3d
4EId
a La
− w4EI
−
a
b Lb
− w4EI
−
b
c Lc
− w4EI
−
c
0
(7)
The 5 × 5 matrix on the left hand side of equation (7) is called a flexibility matrix and is tri-diagonal and symmetric. This equation can be written
symbolically as F m = d. By examining the general form of this expression,
we can write a matrix representation of the three-moment equation for arbitrarily many spans. If a numbering convention is adopted in which support
j lies between span j − 1 and span j, the three non-zero elements in row j of
matrix F are given by
Lj−1
,
EIj−1


L
L
j−1
j
,
= 2
+
EIj−1 EIj
Lj
=
.
EIj
Fj,j−1 =
Fj,j
Fj,j+1
(8)
(9)
(10)
For the case of uniformly distributed loads, row j of vector d is
wj−1 L3j−1 wj L3j
dj = −
−
.
4EIj−1
4EIj

0
(11)
The moments at the supports are computed by solving the system of equations F m = d for the vector m.
CC BY-NC-ND H.P. Gavin













.













6
CEE 201L. – Uncertainty, Design, and Optimization – Duke University – Spring 2009 – H.P. Gavin
Once the internal moments are found, the reactions at the supports can be
computed from static equilibrium.
"r"
"l"
j−1
wj−1
j−1
j
wj−1Lj−1 /2
xL
M j−1
L
PR x R /L
j−1
j−1
Mj−1 /L
j−1
j+1
xR
PR
M j /L
j
w j Lj /2
PL
P x /L
L L
w
j
Mj
j
M j /L j
M j+1
M j+1 /L j
L
j−1
j
xR Mj Mj Mj−1 Mj+1
1
xL
1
+PR −
−
+
+
(12)
Rj = wj−1 Lj−1 + wj Lj +PL
2
2
Lj−1
Lj Lj−1 Lj Lj−1
Lj
where the first two terms on the right hand side correspond to a uniformly
distributed load and the next two terms correspond to interior point loads.
Having computed the reactions and internal moments, we can find the shear
and moment diagrams from equilibrium equations. For example, consider
span j between support j and support j + 1. The internal shear force at
support j in span j is
Vj,j =
Mj − Mj+1 1
xk
− wj L j − Pk ,
Lj
2
Lj
(13)
and the internal shear force at support j + 1 in span j is

Vj+1,j

Mj − Mj+1 1
xk
=
+ wj Lj + Pk 1 −  .
Lj
2
Lj
(14)
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7
Strain Energy in Linear Elastic Solids
The term wj Lj /2 in these two equations corresponds to a uniformly distributed load, Pk is a point load within span j, and xk is the distance from
the right end of span j to the point load Pk .
Beam rotations at the supports may be computed from equations (1), (2),
and (3). The slope of the beam at support j is tan θj . From the second
Moment-Area Theorem,


1 1
1 xk
1
1
tan θj = −
wj L3j + Pk (L2j − x2k ) + Mj Lj + Mj+1 Lj  ,
EIj 24
6 Lj
3
6
(15)
where span j lies between support j and support j + 1. The first term inside
the brackets corresponds to a uniformly distributed load. The second term
inside the brackets corresponds to a point load Pk within span j, located a
distance xk from the right end of span j, (support j + 1).
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8
CEE 201L. – Uncertainty, Design, and Optimization – Duke University – Spring 2009 – H.P. Gavin
These ideas are implemented in the Matlab function three moment.m.
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function [M ,R ,V , xs , M_Diag , V_Diag , d_Diag , D_max ] = three_moment (L ,I ,E ,w ,P , x )
% [M, R, V, xs , M Diag , V Diag , d Diag , D max ] = three moment (L , I , E, w , P, x )
% s o l v e t h e t h r e e −moment e q u a t i o n s f o r a c o n t i n u o u s beam o f N s p a n s .
%
% INPUT :
%
L i s a v e c t o r o f l e n g t h N c o n t a i n i n g t h e l e n g t h s o f each span .
%
I i s a v e c t o r o f l e n g t h N c o n t a i n i n g t h e moments o f i n e r t i a o f each span .
%
E i s a s c a l a r c o n s t a n t f o r t h e modulus o f e l a s t i c i t y .
%
w i s a v e c t o r o f l e n g t h N c o n t a i n i n g t h e uniform l o a d s on each span .
%
P i s a v e c t o r o f p o i n t l o a d s magnitudes
%
x i s a v e c t o r o f p o i n t l o a d l o c a t i o n s , from t h e s t a r t o f t h e f i r s t span
%
% OUTPUT:
%
M
i s a v e c t o r o f l e n g t h N+1 c o n t a i n i n g t h e moments a t each s u p p o r t
%
R
i s a v e c t o r o f l e n g t h N+1 c o n t a i n i n g t h e r e a c t i o n s a t each s u p p o r t
%
V
i s a m a t r i x o f s i z e 2xN+1 c o n t a i n i n g end−s h e a r s o f each span
%
xs
i s a v e c t o r o f t h e x−a x i s f o r t h e s h e a r , moment , and d i s p l p l o t s
%
M Diag i s a v e c t o r o f t h e moment diagram
%
V Diag i s a v e c t o r o f t h e s h e a r diagram
%
d D i a g i s a v e c t o r o f t h e d i s p l a c e m e n t diagram
%
D max i s a v e c t o r o f t h e max a b s o l u t e d i s p l a c e m e n t o f each span
%
% This program assumes t h a t none o f t h e s u p p o r t s a r e moment r e s i s t i n g ,
% t h a t t h e r e a r e no h i n g e s i n t h e beam , t h a t a l l t h e s p a n s a r e made o f
% t h e same m a t e r i a l , and t h a t each s p a n s i s p r i s m a t i c .
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% H. P . Gavin , C i v i l and Environmental E n g i n e e r i n g , Duke U n i v e r s i t y , 3/24/09
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N = length ( L );
% number
o f spans
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nP = length ( x );
% number o f c o n c e n t r a t e d p o i n t l o a d s
span = zeros (1 , nP );
% spans c o n t a i n i n g the p o i n t l o a d s
sumL = cumsum( L );
xL
= zeros (1 , nP );
% d i s t a n c e from p o i n t l o a d t o l e f t r c t n
xR
= zeros (1 , nP );
% d i s t a n c e from p o i n t l o a d t o r i g h t r c t n
f o r i =1: length ( x )
span ( i ) = min( find ( x ( i ) < sumL ) );
end
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f o r k =1: nP
% loop over a l l concentrated point loads
i f span ( k ) == 1
% t h e p o i n t l o a d i s i n t h e f i r s t span
xL ( k ) = x ( k );
else
xL ( k ) = x ( k ) - sumL ( span ( k ) -1);
end
xR ( k ) = sumL ( span ( k )) - x ( k );
end
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F = zeros ( N +1 , N +1);
% i n i t i a l i z e the f l e x i b i l i t y matrix
f o r j =2: N
% c r e a t e t h e f l e x i b i l i t y m a t r i x (8) −(10)
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F (j ,j -1) = L (j -1) / I (j -1);
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F (j , j )
= 2 * (
L (j -1) / I (j -1)
+
L(j) / I(j)
);
CC BY-NC-ND H.P. Gavin
9
Strain Energy in Linear Elastic Solids
56
F (j , j +1) = L ( j ) / I ( j );
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end
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F (1 ,1)
= 1.0;
F ( N +1 , N +1) = 1.0;
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d = zeros ( N +1 ,1);
f o r j =2: N
% c r e a t e t h e r i g h t −hand−s i d e v e c t o r
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l = j -1;
r = j;
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% j −1
% j
i s t h e number o f t h e l e f t
span
i s t h e number o f t h e r i g h t span
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d ( j ) = -w ( l )* L ( l )ˆ3 / (4* I ( l )) - w ( r )* L ( r )ˆ3 / (4* I ( r ));
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f o r k =1: nP
% loop over a l l concentrated point loads
i f span ( k ) == l
% t h e p o i n t l o a d i s i n t h e l e f t span
d ( j ) = d ( j ) - P ( k )* xL ( k )/( L ( l )* I ( l ))*( L ( l )ˆ2 - xL ( k )ˆ2);
end
i f span ( k ) == r
% t h e p o i n t l o a d i s i n t h e r i g h t span
d ( j ) = d ( j ) - P ( k )* xR ( k )/( L ( r )* I ( r ))*( L ( r )ˆ2 - xR ( k )ˆ2);
end
end
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end
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M = ( inv ( F ) * d ) ’;
% compute t h e i n t e r n a l moments
R = zeros (1 , N +1);
f o r j =1: N +1
% build the vector of reaction forces
% j i s t h e r e a c t i o n number
(7)
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l = j -1;
r = j;
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% j −1
% j
i s t h e number o f t h e l e f t
span
i s t h e number o f t h e r i g h t span
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i f j == 1
R ( j ) = w ( r )* L ( r )/2 - M ( j )/ L ( r ) + M ( j +1)/ L ( r );
end
i f j == N +1
R ( j ) = w ( l )* L ( l )/2 - M ( j )/ L ( l ) + M (j -1)/ L ( l );
end
i f j > 1 && j < N +1
R ( j ) = w ( l )* L ( l )/2 + w ( r )* L ( r )/2 ...
- M ( j )/ L ( l ) - M ( j )/ L ( r ) + M (j -1)/ L ( l ) + M ( j +1)/ L ( r );
end
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f o r k =1: nP
% loop over a l l concentrated point loads
i f span ( k ) == l
% t h e p o i n t l o a d i s i n t h e l e f t span
R ( j ) = R ( j ) + P ( k )* xL ( k )/ L ( l );
end
i f span ( k ) == r
% t h e p o i n t l o a d i s i n t h e r i g h t span
R ( j ) = R ( j ) + P ( k )* xR ( k )/ L ( r );
end
end
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end
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slope = zeros (1 , N );
f o r j =1: N
% compute t h e s l o p e s
(15)
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10 CEE 201L. – Uncertainty, Design, and Optimization – Duke University – Spring 2009 – H.P. Gavin
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% j i s t h e span t o t h e r i g h t o f r e a c t i o n j
r = j;
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slope ( j ) =
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w ( r )* L ( r )ˆ3 / 24
+
M ( j +1)* L ( r ) / 6
+
M ( j )* L ( r ) / 3;
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f o r k =1: nP
% loop over a l l concentrated point loads
i f span ( k ) == r
% t h e p o i n t l o a d i s i n t h e r i g h t span
slope ( j ) = slope ( j )+ P ( k )* xR ( k )/ L ( r )*( L ( r )ˆ2 - xR ( k )ˆ2)/6;
end
end
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slope ( j ) = - slope ( j ) / ( E * I ( r ) );
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end
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if (
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abs ( sum( R ) - sum ( w .* L ) - sum( P ) ) < 1e -9 )
disp ( ’ yes ! ’)
% e q u i l i b r i u m c h e c k . . . s h o u l d be c l o s e t o z e r o
end
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% −−−−−−−−−−
s h e a r , moment , s l o p e , and d e f l e c t i o n d a t a and p l o t s
−−−−−−−−−−−
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f o r j =1: N
% x−a x i s d a t a f o r s h e a r , moment , s l o p e , and d e f l e c t i o n diagrams
xs (: , j ) = [ 0: L ( j )/157: L ( j ) ] ’ ;
end
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f o r j =1: N
% j i s t h e span number
Vo = ( M ( j ) - M ( j +1) ) / L ( j ) - w ( j )* L ( j )/2;
% shear at l e f t
V_Diag (: , j ) = Vo + w ( j )* xs (: , j );
f o r k =1: nP
% loop over a l l concentrated point loads
i f span ( k ) == j
% t h e p o i n t l o a d i s i n span t o t h e r i g h t
i1 = find ( xs (: , j ) < xL ( k ));
i2 = find ( xs (: , j ) > xL ( k ));
V_Diag ( i1 , j ) = V_Diag ( i1 , j ) - P ( k )* xR ( k )/ L ( j );
V_Diag ( i2 , j ) = V_Diag ( i2 , j ) + P ( k )*(1 - xR ( k )/ L ( j ));
end
end
M_Diag (: , j ) = M ( j ) + cumtrapz ( - V_Diag (: , j ) ) * xs (2 , j );
s_Diag (: , j ) = cumtrapz ( M_Diag (: , j ) ) * xs (2 , j ) / ( E * I ( j )) + slope ( j ) ;
d_Diag (: , j ) = cumtrapz ( s_Diag (: , j ) ) * xs (2 , j ) ;
end
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157
158
159
160
161
% −−−−−−−−−−−−− d i s p l a y k e y r e s u l t s t o t h e s c r e e n −−−−−−−−−−−−−−
f p r i n t f ( ’ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -\ n ’ );
fprintf (’
Moment
Shear
Deflection \ n ’ );
fprintf (’
Maximum
%12.5 e
%12.5 e
%12.5 e \ n ’ , ...
max(max( M_Diag )) , max(max( - V_Diag )) , max(max( d_Diag )) );
fprintf (’
Minimum
%12.5 e
%12.5 e
%12.5 e \ n ’ , ...
min(min( M_Diag )) , min(min( - V_Diag )) , min(min( d_Diag )) );
f p r i n t f ( ’ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -\ n ’ );
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163
164
165
166
f o r j =1: N
V (1 , j ) = - V_Diag (1 , j );
V (2 , j ) = - V_Diag (158 , j );
end
% j i s t h e span number
% shear force at l e f t
end o f span
% s h e a r f o r c e a t r i g h t end o f span
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168
169
f o r j =2: N
% x−a x i s d a t a f o r s h e a r and moment diagram p l o t s
xs (: , j ) = xs (: , j ) + sumL (j -1);
CC BY-NC-ND H.P. Gavin
Strain Energy in Linear Elastic Solids
170
11
end
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% Plotting
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179
xs
= xs (:);
M_Diag = M_Diag (:);
V_Diag = - V_Diag (:);
s_Diag = s_Diag (:);
d_Diag = d_Diag (:);
z = zeros (1 , length ( xs ));
180
181
D_max
= max(abs( d_Diag ));
182
183
184
185
186
187
f i g u r e (1)
clf
subplot (411)
plot ( xs , z , ’ -k ’ , xs , V_Diag , ’ -b ’ , ’ LineWidth ’ , 2 )
ylabel ( ’ Internal Shear ’)
188
189
190
191
subplot (412)
plot ( xs , z , ’ -k ’ , xs , M_Diag , ’ -b ’ , ’ LineWidth ’ , 2 )
ylabel ( ’ Internal Moment ’)
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195
subplot (413)
plot ( xs , z , ’ -k ’ , xs , s_Diag , ’ -b ’ , ’ LineWidth ’ , 2 )
ylabel ( ’ Slope ’)
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197
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199
subplot (414)
plot ( xs , z , ’ -k ’ , xs , d_Diag , ’ -b ’ , ’ LineWidth ’ , 2 )
ylabel ( ’ Deflection ’)
200
201
% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− three moment .m
CC BY-NC-ND H.P. Gavin
12 CEE 201L. – Uncertainty, Design, and Optimization – Duke University – Spring 2009 – H.P. Gavin
Example
L
I
E
w
P
x
=
=
=
=
=
=
[100
150
150
50];
[ 500 1000 2000 100 ];
1000;
[ 0.10
0.20
0.10
0.30 ];
[ 10
20
40
20
5 ];
[ 110
130
300
330
420 ];
%
%
%
%
%
%
lengths of each span
bending moment of inertia of each span
elastic modulus
uniformly distributed load, each span
concentrated interior point loads
location of point loads from x=0
[M,R,V] = three_moment(L,I,E,w,P,x)
yes!
-----------------------------------------------------------------Moment
Shear
Deflection
Maximum
1.30355e+03
4.89758e+01
1.73453e-01
Minimum
-1.10017e+03
-2.60242e+01
-1.25557e+00
-----------------------------------------------------------------M = 0.0000e+00 -3.0056e+02 -1.1002e+03 -2.7880e+02
0.0000e+00
R = 1.9944
43.0082
73.9732
42.1003
V = 1.9944
-8.0056
35.0026
-24.9974
48.9758
-26.0242
16.0761
-3.9239
3.9239
CC BY-NC-ND H.P. Gavin
13
Internal Moment
Internal Shear
Strain Energy in Linear Elastic Solids
40
30
20
10
0
-10
-20
0
50
100
150
200
250
300
350
400
450
0
50
100
150
200
250
300
350
400
450
0
50
100
150
200
250
300
350
400
450
0
50
100
150
200
250
300
350
400
450
1000
500
0
-500
-1000
Deflection
Slope
0.02
0.015
0.01
0.005
0
-0.005
-0.01
-0.015
-0.02
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-1.2
CC BY-NC-ND H.P. Gavin